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The explanation I have heard is that it is due two newtons third law. The earth pulls on the moon and the moon pulls on the earth with equal and opposite forces for each, which means the force must be the same. The moon pulls the earth toward it, and is also pulled towards the earth. The same is true for the earth, and the resultant force is therefore the same for both. That is my understanding of newtons third law applied to gravity.

But why must this be true? Why when two objects attract each other gravitationally is the resultant force on both objects pulling them together the same? Why must newtons third law be true in this situation. What fundamental explanation for this law can be given to prove that the forces are the same? What would happen if this wasn't the case, and the force was different? Would it break special relativity? Is there a proof for this other than observable evidence and saying that "newtons third law works in other situations so it must work in this situation as well"?

For me, newtons third law makes sense for physical interactions because I can think of molecules and repulsion forces, but what fundamental explanation can be given for action at a distance? (idk if the reason I gave is truly why the third law works for physical interactions, but it makes sense to me and I have heard of no other explanation).

Qmechanic
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Curulian
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2 Answers2

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If two objects didn't have equal and opposite gravitational forces that would break conservation of momentum.

Change in momentum is defined as $\Delta p = F \Delta t$. Conservation of momentum for two objects in a closed system goes as

$p_{1i} + p_{2i} = p_{1f} + p_{2f}$

So therefore $F_1 \Delta t = - F_2\Delta t; F_1 = -F_2$

If, for example, a skydiver gained downward momentum without the earth gaining "upward" momentum (relatively speaking), then the net momentum of the system would not be conserved.

It's reasonable to then ask "why is conservation of momentum true?", and the theory behind that to look at is Noether's theorem, where conservation of momentum follows directly from the homogeneity of space.

Señor O
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I will discuss first the case of a force exerted between two objects by a cable being reeled in.
From there I will move on the the case of gravity.

Let it be granted that $F=ma$ holds good. And I need to make this explicit: let it be granted that $F=ma$ will obtain the same at every location of space, and independent of orientation in space.

Let there be two objects, in space, small enough that any gravitational acceleration is negligable. We name these object 1 and object 2.

(Other than that a large population of objects is present, with gravitational acceleration between these objects negligable, thus the population as a whole provides reference of acceleration for the motions of object 1 and object 2.)

Let a cable be stretched between object 1 and object 2. For the benefit of the thought demonstration we treat that cable as massless. One of the objects starts reeling in that cable.

The following sequence is observed:
State A: first the two are both stationary
They start moving towards each other
When the two come in contact the two stick together and in that final state (state B) they don't have a velocity with respect to State A.

The following is of central importance: the resulting motion does not tell us which object was reeling in the cable.

This is because in space the only leverage you have is your own inertial mass. By contrast: on Earth you can attach yourself to the Earth surface, thus securing yourself to something much more massive than yourself; in space the only leverage that you have is your own inertial mass.

In space the amount of leverage that any object has (for the purpose of reeling in another object) is given by F=ma. The amount of opposition to being reeled in is also given by F=ma

Is the process of reeling in mutual? Or is it one-sided? The point is: that is indistinguishable; on the basis of the resulting motion we cannot eliminate either possibility.

Since we cannot eliminate either possibility we turn that around and make it an element of our theory of motion that no such distinction can be made.

So we have the following:

If it is granted that F=ma will obtain in the same way at every location in space: For motion with respect to the Common Center Of Mass (CCOM) of object 1 and object 2:

$$ m_1 * a_1 = m_2 * -a_2$$

It follows that during the entire time:

$$ m_1 * v_1 = m_2 * -v_2 $$

It follows that with respect to the CCOM the total linear momentum remains zero.

And of course Newton's third law follows logically:

$$ F_{reaction} = -F_{action} $$



Gravity

We infer the existence of gravity from how gravity affects motion. It is the resulting motion that is observable, not gravity itself.

So: what happens if the following question is raised: 'is gravity one-sided, or is it mutual?'. The point is: there is no way to addres that question. Either way the resulting motion is the same.



Discussion:
Philosophy of physics

Quote from physics.stackexchange contributor knzhou
[...] in physics, you can often run derivations in both directions: you can use X to derive Y, and also Y to derive X. That isn't circular reasoning, because the real support for X (or Y) isn't that it can be derived from Y (or X), but that it is supported by some experimental data D. This two-way derivation then tells you that if you have data D supporting X (or Y), then it also supports Y (or X).



So: we need to keep in mind that derivation does not have an intrinsic direction.

Another way of saying this:
In any logical system there is great freedom to exchange axiom and theorem without changing the content of the system.

That said: I have a preference for the sequence as presented above:
The summary:
Take as starting point a combination of $F=ma$, and assertion of uniformity. The assertion of uniformity: $F=ma$ will obtain the same at every location in space.

Those two assertions are sufficient to imply conservation of momentum and Newton's third law.

Cleonis
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